# Effective impedance calculation for noise removal from supply

I am working on single supply Op-Amp amplifier, referring to this post. While calculating the value of the capacitor C2, as to pass all the noise from the supply to the ground. I am unable to understand the logic behind the calculation. If there is noise on the 5V supply, it will pass through R1 and then pass to the ground via C2 capacitor.

1. If R1 & R2 are 10K ohm, then how the effective impedance will act as 5K ohm for the noise. Thus to reduce the noise up to 20 Hz, the value of the capacitor taken should be min 1.6uF. (I am NOT able to understand why parallel combination of 10K ohm resistor should be done, which equals to 5K ohm).

2. If the C3 capacitor is not present will the amplifier will amplify DC voltage as well as ac signal?

3. For calculating the value of C1, as to allow min freq of 5Hz should pass, how is the value of resistance should be calculated? How the resistors R1, R2 & R3 should be involved (I am clueless about it).

4. Same thing for C3, for passing frequencies from 5Hz onward.

I am looking for how to start thinking about the questions, in logical & step-by-step manner.

Original post by Analog Devices is here.

• Olin went to a lot of trouble to explain many of the details. I do agree that Olin doesn't cover every bit of reasoning possible. But a lot of it is covered. Start only with the network of resistors and capacitors feeding the (+) input and tell us what it is about this that you cannot follow, with regard to what Olin writes. Yes, I can see that you say you don't understand. But Olin writes words and you haven't addressed yourself to what Olin writes about this specific network. So I have zero idea what it is about what Olin said that you do NOT understand. So talk a little, please.
– jonk
Nov 29 '19 at 4:33

First of all your confusion stems from thinking about R1, R2, and R3 as if searching for thevenin values, when they are independent of each other. R1 with R2 divide Vcc in half, and if it is known you could say they divide the noise on the 5 volt line in half. R3 should be two or more times the value of R1 so the signal input impedance is high enough not to load down most signal sources.

Your time constant for noise is just T = $$\ 1.1 * R1 * C2\$$. The pass frequency is 1/T = 10 HZ. You are thinking minimal values when many engineers think of the maximum value possible in terms of both cost and space allowed. Your 1.6 uF capacitor would become 4.7 uF (5 HZ) or even 10 uF (1 HZ). Remember that residual noise at C2 affects the DC offset at the op-amp + input, which is amplified by the op-amp gain setting.

You want to go beyond just filtering out noise that is audible, you want to filter out minor low frequency 'thumps' in voltage due to other loads on the 5 volt line. Also you would be staying with know off-the-shelf capacitor values and the closest to 1.6 uF is 2.2 uF.

Analog Devices states this is a more stable single-supply design as R3 isolates C2 from the + input, allowing a stable Vref to be created where R1, R2, R3 and C2 meet. R3 taps into this stable and quiet DC source as bias for the op-amp equal to 1/2 Vcc, or 2.5 volts in this case. R3 also sets the audio input impedance for this amplifier. To avoid noise and hum pickup I would keep R3 under 100 K.

As the value of C2 goes up it acts as a reservoir for stable DC bias current. It is the bias for the op amp so it MUST be stable with essentially zero noise. Combined with adequate power supply decoupling there should be no self induced noise in the op-amps output. In terms of board space and cost a 100 uF for C2 (0.1 HZ) would be optimal for noise roll-off. The R1, R2 and R3 thevenin only has meaning if C2 is not there to dominate as a low-pass filter.

• I think the OP wants to understand why it is that, for example, $R_1$ and $R_2$ should necessarily split the voltage in half. There is a very good reason that isn't just that one might actually want half the supply voltage there. Why is it, for example, that noise or spikes on the Vcc rail are better nipped away by this exact choice? The OP should be told why. The choice isn't just about the need for 1/2 Vcc. It's about isolation and that imbalanced values would lead to closer coupling to Vcc/ground noise/spikes. So midpoint is better if otherwise acceptable.
– jonk
Nov 29 '19 at 5:02
• @jonk. Tough to pin this one down jonk, but C2 is the keystone here.
– user105652
Nov 29 '19 at 5:29
• If you look over many discrete BJT circuits used in older audio amplifiers, you'll find the use of something like this as common. And in cases where half the supply rail is actually not perhaps the best choice (for other reasons.) If either resistance is lower, then that resistance becomes the closer path by which spikes enter. By maximizing the Thevenin resistance (distance), you maximize the isolation. Of course, there are other considerations (the $\tau$, obviously; and leakage currents that may have to pass.) But the whole thing is an interesting subject with many facets to it.
– jonk
Nov 29 '19 at 5:37